I am trying to answer the following:

Implement a function that is given an array of integers and the number of items. It should return the rounded (to the nearest integer) average of the numbers.

This is my code:

#include <stddef.h>
int CalculateAverage (const int array[], size_t size)
{
    
    int i;
    long total;
    float convert;
    int temp = 0;
    int average;
    if (size == 0)
    {
        return 0;
    }
    for (i=0; i<size; i  )
    {
        temp = total;
        total = temp   array[i];
    }
    
    convert = total;
    convert = convert / size;
    average = (convert > 0) ? convert   0.5 : convert - 0.5;
    
    return average;
}

I can't use any functions. When I run the test on our class tutorial app I get (every time with different inputs):

Testing Report:
Running test: CalculateAverage(size=11)  --  Passed
Running test: CalculateAverage(size=13)  --  Failed

Done

I am not sure where is my mistake. I think it is related to negative/positive inputs. any suggestions?

CodePudding user response：

Main issue: Uninitialized object

Below is bad as total is read, but not yet initialized nor assigned. @abelenky

long total;
 ...
    temp = total;

long not that wide

Good that OP is trying to accumulate the sum in a wider type, yet long is not certainly twice as wide as int ^*1. Consider long long or intmax_t. Even those are not certain to be wider, yet commonly are wider.

Wrong rounding

convert 0.5 not needed as prior convert / size already rounded. Floating point math not needed. Conversion to float risks losing precision. Use of double math with float object is curious.

Wrong type

Iterator i should match type of size.

Good use of const and size==0 detection

---

Proposed fix

#include <inttypes.h>
#include <stddef.h>

int CalculateAverage (const int array[], size_t size) {
  if (size == 0) {
    return 0;
  }
  intmax_t sum = 0;
  for (size_t i = 0; i<size; i  ) {
    sum  = array[i];
  }

  // Form a rounded average by subtracting/adding half the size before the division.
  // For handling ties and round to even obliges a little more code.
  sum = (sum < 0) ? sum - size/2 : sum   size/2;
  return (int) (sum / size);
}

---

^*1 The width needed is the width of int plus the width of size_t. To truly handle very extreme averaging, takes additional code.

CodePudding user response：

To add to the comments you could make it more understandable by accumulating the results of division and reminders of division in 2 varaibles like this:

int CalculateAverage(const int array[], size_t size) {
    if (size == 0) return 0;
    int sum_of_divided = 0, sum_of_remainders = 0;
    for (int i = 0; i < size; i  ) {
        sum_of_divided  = array[i] / size;
        sum_of_remainders  = array[i] % size;
    }
    return sum_of_divided   sum_of_remainders / size;
}

This doesn't require floats and is thus not susceptible to floating point precision errors.

CodePudding user response：

Problem solved based on this comment from @abelenky: long total = 0.

CodePudding user response：

//Working example
#include <stdio.h>
#include <stddef.h>
#include <math.h>

void main() {
    
float CalculateAverage (int array[], int arr_size)
{
    int total=0;
    float average=0;
    if (arr_size == 0)
    {
        return 0;
    }
    for (int i=0; i<arr_size; i  )
    {
        total  = array[i];
    }
    printf("Size of array = %d", arr_size);
    printf("\nSum of array = %d", total);
    average = (float) total / arr_size;
    printf("\nActual average = %f", average);
    return average;
}

int numArray[] = {7, 8, 9, 11, 13}; 
int arr_size = sizeof(numArray)/sizeof(numArray[0]);
//You can manually pass the array size if you don't want to use any built-in function
float avg = CalculateAverage(numArray, arr_size);
if(avg>0)
    {
        float tempNum = avg*10;
        float reminder = fmodf(tempNum, 10);
        if(reminder >= 5)
        {
            avg  ;
        }
           printf("\nRounded off average = %d", (int)avg);
    }
    else{
        printf("Average = %lf", avg);
    }
}